The obvious solution is 4 coin, but this impossible since we have at least 100
coins. If we add any multiple of 7 · 11 · 13 = 1001, this does not change the remainder upon
dividing by 7, 11 or 13, so 2002 + 4 is a solution. Alternative solution: let n be the number of
coins. Then n = 7x + 4 = 11y + 4 for some integers x and y so 7x = 11y so x must be divisible
by 11. Thus n = 77z + 4. But also n = 13w + 4 for some w, so now z is divisible by 13, and we
get x = 1001t+4 for some t. To finish with between 1500 and 3000 coins the only possible t we
can take is 2, leading to 2006 coins. It is easy to check that this does indeed give a solution.